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JEE Mathematics 2023 Question with Solution

Let f:R→Rf : \mathbb{R} \to \mathbb{R} be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)−1f(x + y) = f(x) + f(y) - 1 for all x,y∈Rx, y \in \mathbb{R}. If f′(0)=2f'(0) = 2, then ∣f(−2)∣|f(-2)| is equal to:

  • A

    11

  • B

    22

  • C

    33

  • D

    44

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: f(x+y)=f(x)+f(y)−1f(x + y) = f(x) + f(y) - 1 for all x,y∈Rx, y \in \mathbb{R}, and f′(0)=2f'(0) = 2.

Find: ∣f(−2)∣|f(-2)|.

Differentiate the functional equation with respect to yy:

ddy(f(x+y))=ddy(f(x)+f(y)−1)\frac{d}{dy}\bigl(f(x+y)\bigr)=\frac{d}{dy}\bigl(f(x)+f(y)-1\bigr)

So,

f′(x+y)=f′(y)f'(x+y)=f'(y)

for all x,y∈Rx,y \in \mathbb{R}. Hence, f′(x)f'(x) is a constant function.

Let

f′(x)=cf'(x)=c

for all x∈Rx \in \mathbb{R}. Since f′(0)=2f'(0)=2, we get

c=2c=2

Therefore,

f′(x)=2f'(x)=2

Integrating, the general form of the function is

f(x)=2x+Cf(x)=2x+C

where CC is a constant.

Substitute f(x)=2x+Cf(x)=2x+C into the given relation:

f(x+y)=2(x+y)+C=2x+2y+Cf(x+y)=2(x+y)+C=2x+2y+C

and

f(x)+f(y)−1=(2x+C)+(2y+C)−1=2x+2y+2C−1f(x)+f(y)-1=(2x+C)+(2y+C)-1=2x+2y+2C-1

Equating both sides,

2x+2y+C=2x+2y+2C−12x+2y+C=2x+2y+2C-1

So,

C=2C−1C=2C-1

which gives

C=1C=1

Thus,

f(x)=2x+1f(x)=2x+1

Now evaluate at x=−2x=-2:

f(−2)=2(−2)+1=−4+1=−3f(-2)=2(-2)+1=-4+1=-3

Hence,

∣f(−2)∣=3|f(-2)|=3

Therefore, the correct option is C.

Common mistakes

  • Differentiating with respect to the wrong variable and concluding an incorrect relation. Here, differentiating with respect to yy gives f′(x+y)=f′(y)f'(x+y)=f'(y), which shows the derivative is constant; do not treat f(x)f(x) as varying with yy.

  • Finding f′(x)=2f'(x)=2 and then writing f(x)=2xf(x)=2x directly. This misses the constant of integration. You must write f(x)=2x+Cf(x)=2x+C and then use the functional equation to determine CC.

  • Using the functional equation incorrectly while substituting f(x)=2x+Cf(x)=2x+C. Both sides must be expanded carefully; otherwise the equation for CC is missed. Compare coefficients and constants exactly to get C=1C=1.

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